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\begin{document}

\title{高等代数二}
\subtitle{8-2-正交基 }
%\institute{上海立信会计金融学院}
%\author{王立庆}
\author{{\ppr LQW}}
\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
%\date{{\ppr 2023年3月9日} }

\maketitle

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{8.2.i. 作业：星期天晚上十点半之前在网络教学平台提交 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{enumerate}
\item   整理课堂笔记，补充没写完的计算或证明。
\item   习题(8.2)\#1,2,5,7,10, 抄写题目。
\end{enumerate}

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\begin{frame}{8.2.ii. 目录 }

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\begin{enumerate}

\item[8.2.1.] 正交组的定义
\item[8.2.5.] 定理8.2.1. 正交组是线性无关的向量组
\item[8.2.6.] 规范正交基的定义
\item[8.2.8.] 定理8.2.2. 斯密特正交化方法
\item[8.2.9.] 定理8.2.3. 正交基的规范化
\item[8.2.11.] 正交补空间的定义
\item[8.2.13.] 定理8.2.4. 正交补空间是余子空间
\item[8.2.14.] 正射影的定义
\item[8.2.16.] 定理8.2.5. 正射影的极小性质
\item[8.2.18.] 正交矩阵的定义
\item[8.2.20.] 定理8.2.6. 规范正交基之间的过渡矩阵是正交矩阵

\end{enumerate}

\end{frame}

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\begin{frame}{8.2.iii. 课堂讲解重点 }

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\begin{enumerate}
\item  规范正交基的概念 
\item  格莱姆-斯密特正交化方法
\item  正交补空间的概念
\item  正交矩阵
\end{enumerate}

\end{frame}



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\begin{frame}{8.2.1. 正交组的定义 }

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：欧氏空间中的一个向量组，什么时候称为正交组？}

\item  解答：设有非零向量组 $\Phi = \{\alpha_1,\cdots,\alpha_m\}$，如果其中的任意两个向量都是相互正交的，
即对任意 $1\le ,i,j\le m$, $i\neq j$, 都有 $\langle \alpha_i,\alpha_j \rangle =0$, 
那么称这个向量组是一个正交组。

\item  注：向量组 $\Phi = \{\alpha_1,\cdots,\alpha_m\}$ 是正交组，当且仅当它的格拉姆矩阵
{\footnotesize 
$$
\langle \Phi, \Phi \rangle = \begin{pmatrix}
\langle \alpha_1, \alpha_1 \rangle & \langle \alpha_1, \alpha_2 \rangle &\cdots & \langle \alpha_1, \alpha_m \rangle  \\ 
\langle \alpha_2, \alpha_1 \rangle & \langle \alpha_2, \alpha_2 \rangle &\cdots & \langle \alpha_2, \alpha_m \rangle  \\ 
\vdots & \vdots & & \vdots \\ 
\langle \alpha_m, \alpha_1 \rangle & \langle \alpha_m, \alpha_2 \rangle &\cdots & \langle \alpha_m, \alpha_m \rangle  \\ 
\end{pmatrix}
$$
}
是一个对角矩阵，而且对角线元素都不为零。


\end{itemize}

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\begin{frame}{8.2.2. 规范正交组的定义}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}问题：欧氏空间中的一个向量组，什么时候称为规范正交组？}

\item  解答：设有非零向量组 $\Phi = \{\alpha_1,\cdots,\alpha_m\}$, 如果下述条件成立，
\begin{enumerate}
\item  向量组 $\Phi$ 中的任意两个向量都是相互正交的，
\item  向量组 $\Phi$ 中的每个向量都是单位向量，
\end{enumerate}
那么称这是一个规范正交组。

\item  注：向量组 $\Phi= \{\alpha_1,\cdots,\alpha_m\}$ 是规范正交组，当且仅当它的格拉姆矩阵 $\langle \Phi,\Phi \rangle = E_m$ 为单位矩阵，即 
$$
\langle \alpha_i,\alpha_j \rangle =\delta_{ij} 
=\left\{\begin{array}{rcl}
1, & \text{ 当 } i=j, \\ 
0, & \text{ 当 } i\neq j.
\end{array}\right.
$$

\end{itemize}

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\begin{frame}{8.2.3. 例子 }

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：验证下述向量组是 $\mathbb{R}^3$ 中的规范正交组，在立体空间中画出图像，
$$\alpha_1=(0,1,0), \alpha_2=(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}), \alpha_3=(\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}}).$$
}

\item  解答：直接计算这三个向量的两两之间的内积，可得 
{\footnotesize 
$$
\begin{pmatrix}
\langle \alpha_1, \alpha_1 \rangle & \langle \alpha_1, \alpha_2 \rangle & \langle \alpha_1, \alpha_3 \rangle  \\ 
\langle \alpha_2, \alpha_1 \rangle & \langle \alpha_2, \alpha_2 \rangle & \langle \alpha_2, \alpha_3 \rangle  \\ 
\langle \alpha_3, \alpha_1 \rangle & \langle \alpha_3, \alpha_2 \rangle & \langle \alpha_3, \alpha_3 \rangle  \\ 
\end{pmatrix}
= \begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}. 
$$
}
因此向量组 $\{\alpha_1, \alpha_2, \alpha_3\}$ 是规范正交组。


\end{itemize}

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：设 $V=C[0,2\pi]$ 是区间 $[0,2\pi]$ 上的实值连续函数全体组成的向量空间。设内积定义为
$$\langle f,g \rangle =\int_0^{2\pi} f(x)g(x)dx.$$
证明下述向量组是 $V$ 中的一个正交组，
$$1,\cos(x), \sin(x), \cos(2x), \sin(2x), \cdots \cos(nx),\sin(nx),\cdots $$
}

\item  思路：直接计算这些向量的两两之间的内积。

\end{itemize}

\end{frame}

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\begin{frame}{8.2.5. 定理8.2.1.}

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%每页详细内容

\begin{itemize}

\item  {\color{red}命题：欧氏空间中的正交组一定是线性无关的。也就是说，如果欧氏空间 $V$ 中的一个非零向量组 $\Phi=\{\alpha_1,\cdots,\alpha_n\}$ 中的向量是两两正交的，那么这个向量组 $\Phi$ 是线性无关的。
}

\item  证明：
\begin{enumerate}
\item  设有实数 $k_1,\cdots,k_n$ 使得 $k_1\alpha_1 + \cdots + k_n\alpha_n = \theta$.  
\item  两边与 $\alpha_1$ 计算内积，可得 
$\langle k_1\alpha_1 + \cdots + k_n\alpha_n, \alpha_1 \rangle = \langle \theta, \alpha_1 \rangle$. 
\item  根据内积的线性，与两两正交的条件，可得  $k_1\langle \alpha_1, \alpha_1 \rangle =0$. 
\item  因为 $\alpha_1$ 不是零向量，所以 $\langle \alpha_1, \alpha_1 \rangle \neq 0$, 所以 $k_1=0$. 
\item  类似地，可证 $k_2=0, \cdots, k_n=0$. 

\end{enumerate}


\end{itemize}

\end{frame}

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\begin{frame}{8.2.6. 规范正交基的定义}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}问题：欧氏空间中的一个向量组，什么时候称为正交基、规范正交基？}

\item  解答：
\begin{enumerate}
\item  向量组 $\Phi$ 是一个正交基的定义：
\begin{enumerate}
%\item  欧氏空间 $V$ 的一个正交组如果能够张成整个空间，那么就称为一个正交基。
\item  向量组 $\Phi$ 是一个基。
\item  向量组 $\Phi$ 中的任意两个向量都是相互正交的。
\end{enumerate}

\item  向量组 $\Phi$ 是一个规范正交基的定义：
\begin{enumerate}
%\item  一个规范正交组如果能够张成整个空间，那么久称为一个规范正交基。
\item  向量组 $\Phi$ 是一个基。
\item  向量组 $\Phi$ 中的任意两个向量都是相互正交的。
\item  向量组 $\Phi$ 中的每个向量都是单位向量。
\end{enumerate}

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{8.2.7. 例子 }

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：证明欧氏空间 $\mathbb{R}^n$ 的标准基是一个规范正交基。}

\item  解答：以 $n=4$ 为例，
\begin{eqnarray*}
\Phi &=& \{ \varepsilon_1, \varepsilon_2, \varepsilon_3, \varepsilon_4 \} \\ 
&=& \{ (1,0,0,0), \, (0,1,0,0), \, (0,0,0,1), \, (0,0,0,1) \}.
\end{eqnarray*}

验证格拉姆矩阵 $\langle \Phi, \Phi \rangle = E_4$, 即
{\footnotesize 
$$
\begin{pmatrix}
\langle \alpha_1, \alpha_1 \rangle & \langle \alpha_1, \alpha_2 \rangle & \langle \alpha_1, \alpha_3 \rangle & \langle \alpha_1, \alpha_4 \rangle \\ 
\langle \alpha_2, \alpha_1 \rangle & \langle \alpha_2, \alpha_2 \rangle & \langle \alpha_2, \alpha_3 \rangle & \langle \alpha_2, \alpha_4 \rangle\\ 
\langle \alpha_3, \alpha_1 \rangle & \langle \alpha_3, \alpha_2 \rangle & \langle \alpha_3, \alpha_3 \rangle & \langle \alpha_3, \alpha_4 \rangle\\ 
\langle \alpha_4, \alpha_1 \rangle & \langle \alpha_4, \alpha_2 \rangle & \langle \alpha_4, \alpha_3 \rangle & \langle \alpha_4, \alpha_4 \rangle\\ 
\end{pmatrix}
= \begin{pmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1  \end{pmatrix}. 
$$
}

\end{itemize}

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\begin{frame}{8.2.8. 定理8.2.2. 斯密特正交化方法}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  {\color{red}定理：设有欧氏空间 $V$ 中的线性无关的一个向量组 $\{\alpha_1,\alpha_2,\cdots,\alpha_m\}$. 那么一定可以找到一个正交向量组 $\{\beta_1,\beta_2,\cdots,\beta_m\}$ 满足下述条件：
{\footnotesize 
\begin{eqnarray*}
L(\alpha_1) &=& L(\beta_1), \\ 
L(\alpha_1,\alpha_2) &=& L(\beta_1,\beta_2), \\
\cdots && \cdots \\ 
L(\alpha_1,\alpha_2,\cdots,\alpha_m) &=& L(\beta_1,\beta_2,\cdots,\beta_m). 
\end{eqnarray*}
}
}
\vspace{-0.5cm}
\item  证明：以 $m=3$ 为例。
\begin{enumerate}
\item   取 $\beta_1=\alpha_1$. 
\item   取 $\beta_2=\alpha_2- k_1\beta_1$, 适当的 $k_1$ 使得 $\langle \beta_2,\beta_1 \rangle =0$. 
\item   取 $\beta_3=\alpha_3- \ell_1\beta_1-\ell_2\beta_2$, 
适当的 $\ell_1,\ell_2$ 使得 $\langle \beta_3,\beta_1 \rangle =0$, $\langle \beta_3,\beta_2 \rangle =0$.  
%\item   
\end{enumerate}

\end{itemize}

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\begin{frame}{8.2.9. 定理8.2.3. 正交基的规范化}

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%每页详细内容

\begin{itemize}
\item   {\color{red}命题：设 $V$ 是一个有限维欧氏空间。}
\begin{enumerate}
\item  {\color{red}从一个正交基 $\{\beta_1,\beta_2,\cdots,\beta_n\}$ 出发可以得到一个规范正交基。} 
\item  {\color{red}任意一个有限维的欧氏空间都存在规范正交基。} 
\end{enumerate}

\item  证明：
\begin{enumerate}
\item   将正交基的每个向量的长度都变成1. 以 $n=3$ 为例。
\begin{enumerate}
\item   取 $\gamma_1=c_1\beta_1$, 其中 $c_1=\lvert \beta_1 \rvert ^{-1}$.  
\item   取 $\gamma_2=c_2\beta_2$, 其中 $c_2=\lvert \beta_2 \rvert ^{-1}$.  
\item   取 $\gamma_3=c_3\beta_3$, 其中 $c_3=\lvert \beta_3 \rvert ^{-1}$.  
\end{enumerate}

\item   有限维的欧氏空间存在一个基。由此出发，通过斯密特正交化和规范化，可以得到一个规范正交基。

\end{enumerate}

\end{itemize}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}问题：考虑欧氏空间 $\mathbb{R}^3$ 中的下述向量组，
$$\alpha_1=(1,1,1), \alpha_2=(0,1,2), \alpha_3=(2,0,3).$$
用斯密特正交化方法，将其化成正交基。然后再化成规范正交基。
}

\item  解答：
\begin{enumerate}
\item  先正交化：
\begin{enumerate}
\item  $\beta_1 = \alpha_1$. 
\item  $\beta_2 = \alpha_2 - k_1 \beta_1$. 
\item  $\beta_3 = \alpha_3 - m_1 \beta_1 - m_2\beta_2$. 
\end{enumerate}


\item  再规范化：
\begin{enumerate}
\item  $\gamma_1 = c_1\beta_1$. 
\item  $\gamma_2 = c_2\beta_2$. 
\item  $\gamma_3 = c_3\beta_3$. 
\end{enumerate}

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{itemize}

\item  {\color{red}问题：设 $V$ 是一个欧氏空间，设 $W$ 是一个向量子空间。记 
$$W^{\,\perp} = \{\alpha\in V\mid  \forall \beta\in W:\,\, \langle \alpha,\beta\rangle =0\}.$$
验证这样定义的子集 $W^{\,\perp}$ 是 $V$ 的一个向量子空间。
}

\item  解答：
\begin{enumerate}
\item  记 $\theta$ 是 $V$ 的零向量。
因为对任意 $\beta\in W$, 有 $\langle \theta,\beta\rangle =0$, 所以 $\theta\in W^{\,\perp}$. 

\item  设 $\alpha_1, \alpha_2\in W^{\,\perp}$. 
则对任意 $\beta\in W$, 有 $\langle \alpha_1,\beta\rangle =0$, $\langle \alpha_2,\beta\rangle =0$. 
所以有 $\langle \alpha_1+\alpha_2,\beta\rangle =0$. 
所以 $\alpha_1+\alpha_2 \in W^{\,\perp}$. 

\item  设  $\alpha \in W^{\,\perp}$. 设 $k\in \mathbb{R}$. 
则对任意 $\beta\in W$, 有 $\langle \alpha,\beta\rangle =0$. 
所以有 $\langle k\alpha,\beta\rangle = k\langle \alpha,\beta\rangle =0$. 
所以 $k\alpha\in W^{\,\perp}$. 

\end{enumerate}

\end{itemize}

\end{frame}

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：在欧氏空间 $\mathbb{R}^3$ 中，计算 下述子空间的正交补空间，
$$W=\{(x,y,0)\mid x,y\in\mathbb{R}\}.$$
}

\item  解答：$$W^{\,\perp} = \{(0,0,z)\mid z\in\mathbb{R}\}. $$

\end{itemize}

\end{frame}

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%每页详细内容

\begin{itemize}
\item  {\color{red}定理：设 $V$ 是一个欧氏空间，设 $W$ 是 $V$ 的一个有限维子空间。证明 $W^{\,\perp}$ 是 $W$ 的余子空间，也就是说，$V=W+ W^{\,\perp}$ 而且 $W\cap W^{\,\perp}=\{\theta\}$. 简写就是 $V=W\oplus W^{\,\perp}$.
}

\item  证明：
\begin{enumerate}
\item  因为 $W$ 是有限维的，所以存在 $W$ 的规范正交基 $\{\gamma_1, \gamma_2, \cdots, \alpha_m\}$. 
\item  设任意 $\alpha\in V$, 记 $\eta=\langle \alpha,\gamma_1 \rangle \gamma_1 + \langle \alpha,\gamma_2 \rangle \gamma_2 +\cdots + \langle \alpha,\gamma_m \rangle \gamma_m$, 则 $\eta\in W$. 
\item  记 $\beta = \alpha - \eta$. 
\item  对 $k=1,2,\cdots, m$, 因为 $\langle \alpha,\gamma_k \rangle = \langle \eta,\gamma_k \rangle$, 所以 
$\langle \beta, \gamma_k \rangle =0$. 
\item  所以 $\beta\in W^{\,\perp}$. 
\item  因为 $\eta\in W, \beta\in W^{\,\perp}$, 所以 $\alpha = \eta + \beta \in W+W^{\,\perp}$. 
\item  如果 $\xi\in W\cap W^{\,\perp}$, 那么 $\langle \xi,\xi \rangle = 0$. 由正定性可得 $\xi=\theta$ 是零向量。
\item  所以 $V=W\oplus W^{\,\perp}$. 

\end{enumerate}

%\item  思考：如果这个子空间 $W$ 不是有限维的，会有什么问题吗？

\end{itemize}

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：什么是正射影？}

\item  解答：设 $W$ 是欧氏空间 $V$ 的一个有限维子空间。对任意向量 $\alpha\in V$, 根据 $V=W\oplus W^{\,\perp}$, 可知存在唯一的向量 $\alpha_1\in W$ 和 $\alpha_2\in W^{\,\perp}$, 使得 $\alpha=\alpha_1+\alpha_2$. 这时称 $\alpha_1$ 是向量 $\alpha$ 在子空间 $W$ 中的正射影。

%\begin{center}
\includegraphics[height=0.35\textheight,width=0.45\textwidth]{pic/orth-basis-0.png}
%\end{center}

\item  设 $\{\gamma_1, \gamma_2, \cdots, \alpha_m\}$ 是 $W$ 的规范正交基。则向量 $\alpha$ 在 $W$ 中的正射影是 
$$\alpha_1 = \langle \alpha,\gamma_1 \rangle \gamma_1 + \langle \alpha,\gamma_2 \rangle \gamma_2 +\cdots + \langle \alpha,\gamma_m \rangle \gamma_m. $$

\end{itemize}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}问题：在欧氏空间 $\mathbb{R}^3$ 中，计算向量 $\alpha=(1,2,3)$ 下述子空间的正射影，
$$W=\{(x,x,0)\mid x\in\mathbb{R}\}.$$
}

\item  思路：设所求正射影为 $\alpha_1 = (t,t,0)$, 其中 $t$ 是待定系数。
则从 $\alpha - \alpha_1 \perp W$ 可得一个方程。从中求出 $t$. 

%\begin{center}
\includegraphics[height=0.40\textheight,width=0.4\textwidth]{pic/orth-basis-1.png}
%\end{center}

\end{itemize}

\end{frame}


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\begin{itemize}

\item  {\color{red}定理：设 $W$ 是欧氏空间 $V$ 的一个有限维子空间。对任意向量 $\alpha\in V$, 设 $\alpha_1$ 是向量 $\alpha$ 在子空间 $W$ 中的正射影。设 $\beta$ 是 $W$ 中的任意一个向量。证明
$$\lvert \alpha-\alpha_1 \rvert \le \lvert \alpha-\beta \rvert.$$
}

\item  思路：
\begin{enumerate}
\item  将 $\alpha-\beta$ 写成 $\alpha-\alpha_1 + \alpha_1-\beta$. 
\item  证明向量 $\alpha-\alpha_1$ 与向量 $\alpha_1-\beta$ 正交。
\item  使用勾股定理。

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{8.2.17. 例子 }

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\begin{itemize}

\item  {\color{red}问题：设 $V=C[0,2\pi]$ 是区间 $[0,2\pi]$ 上的实值连续函数全体组成的向量空间。
设内积定义为 $$\langle f,g \rangle =\int_0^{2\pi} f(x)g(x)dx.$$
设子空间 $W$ 由下述向量组线性张成，$$\{1,\cos(x), \sin(x), \cos(2x), \sin(2x),\cos(3x),\sin(3x)\}.$$
求向量 $f(x)=x$ 在这个子空间的正射影。
}

\item  思路：先从子空间 $W$ 的生成向量组出发，得到一个规范正交组。然后按公式直接写出正射影。


\end{itemize}

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\begin{frame}{8.2.18. 正交矩阵的定义}

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\begin{itemize}

\item  {\color{red}问题：什么是正交矩阵？}

\item  解答：设 $A$ 是一个 $n$ 阶实数矩阵。 如果有  
$$A^tA=E_n,\,\, AA^t=E_n,$$ 
这里 $E_n$ 是 $n$ 阶的单位阵，那么称矩阵 $A$ 是一个正交矩阵。

\item  注：实数矩阵 $A$ 是正交矩阵当且仅当 $A^{-1}=A^t$. 

\end{itemize}

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\begin{frame}{8.2.19. 例子 }

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\begin{itemize}

\item  {\color{red}问题：验证下述矩阵是正交矩阵：
{\footnotesize 
$$A=\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}.$$ 
}

}

\item  解答：验证正交矩阵的定义，
{\footnotesize 
$$A^tA = \begin{pmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{pmatrix}
\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}
=\begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}.
$$ 
}

\vfill 
\item  注：还有其它形式的二阶的正交矩阵吗？

\end{itemize}

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\begin{frame}{8.2.20. 定理8.2.6. }

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%每页详细内容

\begin{itemize}
\item  {\color{red}定理：欧氏空间的两个规范正交基之间的过渡矩阵是正交矩阵。} 

\item  证明：
\begin{enumerate}
\item  设 $\Phi$ 是一个规范正交基，则有 $\langle \Phi, \Phi \rangle = E$. 
\item  设 $\Psi$ 是另一个规范正交基，则有 $\langle \Psi, \Psi \rangle = E$. 
\item  设从 $\Phi$ 到 $\Psi$ 的过渡矩阵是 $T$, 即 $\Psi = \Phi \cdot T$. 
\item  从上面三个等式得出 $T^tT=E$. 
\end{enumerate}

\end{itemize}

\end{frame}


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\begin{frame}{8.2.21. 欧氏空间之间的同构 }

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：什么是欧氏空间之间的同构？}

\item  解答：设 $V$ 和 $W$ 是两个欧氏空间，称一个映射 $\sigma:V\to W$ 是同构，如果下述两个条件成立：
\begin{enumerate}
\item 映射 $\sigma$ 是实向量空间之间的一个同构。
\item 映射 $\sigma$ 保持内积运算，即对任意 $\alpha,\beta\in V$, 都有 
$$\langle \sigma(\alpha),\sigma(\beta) \rangle = \langle \alpha,\beta \rangle. $$

\end{enumerate}



%\item  问题：上述等号两边的内积，分别是哪个欧氏空间里的？
%\item  解答：%这里等号左边是 $V'$ 中的内积，等号右边是 $V$ 中的内积。




\end{itemize}

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\begin{frame}{8.2.22. 例子 }

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：设 $V$ 是一个 $n$ 维欧氏空间。}
\begin{enumerate}
\item  {\color{red}构造欧氏空间之间的一个同构：$\sigma: V\to \mathbb{R}^n.$ }
\item  {\color{red}在 $V$ 和 $\mathbb{R}^n$ 之间，能构造多少个同构？ }
\end{enumerate}

\item  思路：
\begin{enumerate}
\item  取 $V$ 的一个规范正交基，将其对应到 $\mathbb{R}^n$ 的标准基。
\item  因为 $V$ 有无穷多个规范正交基，所以可以构造无穷多个同构。

%两个规范正交基之间的过渡矩阵是正交矩阵。
\end{enumerate}


\end{itemize}

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\begin{frame}{习题(8.2)\#1 }

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：对下述向量组进行正交化方法，求出 $\mathbb{R}^4$ 的一个规范正交基，
\begin{eqnarray*}
\alpha_1 &=& (0,2,1,0), \\ 
\alpha_2 &=& (1,-1,0,0), \\ 
\alpha_3 &=& (1,2,0,-1), \\ 
\alpha_4 &=& (1,0,0,1).
\end{eqnarray*}
 
}

\item  思路：按照斯密特正交化方法进行计算。



\end{itemize}

\end{frame}

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{习题(8.2)\#2 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}问题：在欧氏空间 $C[-1,1]$ 里，对于线性无关的向量组 $\{1,x,x^2,x^3\}$ 施行正交化方法，求出一个规范正交组。

}

\item  思路：按照斯密特正交化方法进行计算。



\end{itemize}

\end{frame}

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{习题(8.2)\#5 }

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：设向量组 $\{\alpha_1,\alpha_2,\cdots,\alpha_m\}$ 是欧氏空间 $V$ 的一个规范正交组。
证明：对于任意 $\xi\in V$, 成立不等式 
%$$ \sum\limits_{i=1}^m  \langle \xi,\alpha_i \rangle ^2 \le \lvert \xi \rvert ^2. $$
$$ \langle \xi,\alpha_1 \rangle ^2 + \langle \xi,\alpha_2 \rangle ^2 +\cdots +\langle \xi,\alpha_m \rangle ^2 \le \lvert \xi \rvert ^2. $$
}

\item  思路：考虑向量 $\xi$ 在子空间 $L(\alpha_1,\alpha_2,\cdots,\alpha_m)$ 的正射影。



\end{itemize}

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{习题(8.2)\#7 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  {\color{red}问题：证明 $\mathbb{R}^3$ 中的向量 $\xi=(x_0,y_0,z_0)$ 到平面 
$$W=\{ (x,y,z)\in \mathbb{R}^3 \mid ax+by+cz=0 \}$$
的最短距离等于 $$\frac{\lvert ax_0+by_0+cz_0 \rvert }{\sqrt{a^2+b^2+c^2}}. $$
}

\item  思路：计算向量 $\xi$ 在子空间 $W$ 的正射影。


\end{itemize}

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{习题(8.2)\#10 }

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%每页详细内容

\begin{itemize}

\item  {\color{red}问题：设 $A$ 是一个正交矩阵。证明：}
\begin{enumerate}
\item  {\color{red}矩阵 $A$ 的行列式的值等于$\pm 1$. }
\item  {\color{red}矩阵 $A$ 的特征值的模长等于$1$. }
\item  {\color{red}如果 $\lambda$ 是矩阵 $A$ 的特征值，那么 $1/\lambda$ 也是矩阵 $A$ 的特征值。 }
\item  {\color{red}矩阵 $A$ 的伴随矩阵 $A^*$ 也是正交矩阵。}
\end{enumerate}

\item  思路：
\begin{enumerate}
\item  按定义，如果 $A$ 是正交矩阵，那么 $AA^t=E$. 
\item  设在复数范围内有 $AX=\lambda X$. 考虑转置共轭，得到 $\bar{X}^tA^t = \bar{\lambda}\bar{X}^t$. 
\item  设在复数范围内有 $AX=\lambda X$. 考虑共轭，得到 $A\bar{X}=\bar{\lambda} \bar{X}$.
\item  伴随矩阵与逆阵的关系是 $A^{-1}=(\det(A))^{-1}A^*$. 
\end{enumerate}


\end{itemize}

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